This is our solution and implementation to problem #84 on Project Euler.
Our code is written in TypeScript, a language which is built on-top of JavaScript and transpiles to it. We've included the problem statement, our code (which is commented for greater clarity), our video which outlines our analysis and implementation approach, and the solution + how long it took to calculate it.
Note: the code and contents here might be slightly different than what is in the video. We've made some improvements to some of the code since recording.
If you would like to view the original problem and solve it, please visit: Monopoly Odds on Project Euler. If you're having trouble solving this problem, or are just curious to see how others have solved it, feel free to take a look, but please put solid effort into solving this before viewing the actual solution to the problem.
Problem Statement
In the game, Monopoly, the standard board is set up in the following way:
A player starts on the GO square and adds the scores on two 6-sided dice to determine the number of squares they advance in a clockwise direction. Without any further rules we would expect to visit each square with equal probability: 2.5%. However, landing on G2J (Go To Jail), CC (community chest), and CH (chance) changes this distribution.
In addition to G2J, and one card from each of CC and CH, that orders the player to go directly to jail, if a player rolls three consecutive doubles, they do not advance the result of their 3rd roll. Instead they proceed directly to jail.
At the beginning of the game, the CC and CH cards are shuffled. When a player lands on CC or CH they take a card from the top of the respective pile and, after following the instructions, it is returned to the bottom of the pile. There are sixteen cards in each pile, but for the purpose of this problem we are only concerned with cards that order a movement; any instruction not concerned with movement will be ignored and the player will remain on the CC/CH square.
- Community Chest (2/16 cards):
- Advance to GO
- Go to JAIL
- Chance (10/16 cards):
- Advance to GO
- Go to JAIL
- Go to C1
- Go to E3
- Go to H2
- Go to R1
- Go to next R (railway company)
- Go to next R
- Go to next U (utility company)
- Go back 3 squares.
The heart of this problem concerns the likelihood of visiting a particular square. That is, the probability of finishing at that square after a roll. For this reason it should be clear that, with the exception of G2J for which the probability of finishing on it is zero, the CH squares will have the lowest probabilities, as 5/8 request a movement to another square, and it is the final square that the player finishes at on each roll that we are interested in. We shall make no distinction between "Just Visiting" and being sent to JAIL, and we shall also ignore the rule about requiring a double to "get out of jail", assuming that they pay to get out on their next turn.
By starting at GO and numbering the squares sequentially from 00 to 39 we can concatenate these two-digit numbers to produce strings that correspond with sets of squares.
Statistically it can be shown that the three most popular squares, in order, are JAIL (6.24%) = Square 10, E3 (3.18%) = Square 24, and GO (3.09%) = Square 00. So these three most popular squares can be listed with the six-digit modal string: 102400.
If, instead of using two 6-sided dice, two 4-sided dice are used, find the six-digit modal string.
Our Solution
Our solution is given in the TypeScript files below. This solution uses more than one code file. Some solutions use utilities which were created and enhanced while working on this and previous Project Euler problems. Some code in the utilities files might not be used in this particular problem.
Results
This implementation found the solution in 2528ms.
If you would like to view the answer, click below to reveal. Please consider reviewing the implementation and trying to code your own solution before viewing the answer.
View Answer
The answer is 101524.
All of our solutions are hosted on GitHub. The code on this page was pulled from the repo and the solution and execution time were calculated based on that code.